Why does forwarding the name of a function to cout return 1?

Asked Oct 13 '14 at 02:45

Active Oct 13 '14 at 02:45

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In C++, when you write this inside of main

cout << main << endl;

the output is

Can anyone explain why this is? It does it for any function created from my experience, but since main is the default function in most programs, that is the one I used in my example.

asked Oct 13 '14 at 02:45

avbor

2

The only suitiable overload of `operator<<` is the one taking `bool`. That's the one being called. `1` just means "non-NULL pointer" here. – Igor Tandetnik Oct 13 '14 at 02:46
It's possible to write your own overload, that's what makes `std::endl` work. – o11c Oct 13 '14 at 02:47
@IgorTandetnik if you could expand on that it would make a great answer. – Mark Ransom Oct 13 '14 at 02:49

Why does forwarding the name of a function to cout return 1?

0 Answers0